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I am attempting to manually encrypt a plaintext message (message = MI) using RSA.

Manual Encryption of Plaintext Message

I receive an answer of: 33,264 and 21,164.

When I enter the same plaintext into CrypTool to confirm that my calculations were correct, I receive a different answer:

CrypTool Screenshot

What am I doing incorrect? How can I obtain the same result as CrypTool?

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  • $\begingroup$ I checked it with version 1.4.42: When using the default setting of CrypTool 1 it works well and shows the results you expected. This fits with the answer of Peregrinus. $\endgroup$
    – Yves
    Jan 20, 2022 at 14:15

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"The Input text will be separated into segments of Size 2 (the symbol '$' is used as a separator)."

Emphasis mine. It's not encrypting one character at a time, it's encoding "MI" to the number $360$, then computing $360^{11}\mod40741=11807$. It's doing this because you changed some of the "Alphabet and number system options..." settings.

Of note, this is demonstrating textbook RSA, not real RSA as used in practice. Real RSA uses special "padding" schemes to prevent several possible attacks. Textbook RSA is one of the several steps in real RSA encryption (or more commonly signatures).

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